The default hashing algorithm is platform dependent.
Note that it is allowed for two objects to have identical hash
codes (o1.hashCode.equals(o2.hashCode)) yet not be
equal (o1.equals(o2) returns false). A
degenerate implementation could always return 0.
However, it is required that if two objects are equal
(o1.equals(o2) returns true) that they
have identical hash codes
(o1.hashCode.equals(o2.hashCode)). Therefore, when
overriding this method, be sure to verify that the behavior is
consistent with the equals method.

It is reflexive: for any instance x of type Any,
x.equals(x) should return true.

It is symmetric: for any instances x and y of type
Any, x.equals(y) should return true if and only
if y.equals(x) returns true.

It is transitive: for any instances
x, y, and z of type AnyRef
if x.equals(y) returns true and
y.equals(z) returns
true, then x.equals(z) should return true.

If you override this method, you should verify that
your implementation remains an equivalence relation.
Additionally, when overriding this method it is often necessary to
override hashCode to ensure that objects that are
"equal" (o1.equals(o2) returns true)
hash to the same Int
(o1.hashCode.equals(o2.hashCode)).

Parameters

arg0 - the object to compare against this object for equality.

Returns

true if the receiver object is equivalent to the argument; false otherwise.